In hexogonal systems of crystals, a frequently encountered arrangement of atoms is described as a hexagonal prism. Here, the top and bottom of the cell are regular hexongonas and three atoms are sandwiched inbetween them. A space filling model of this structure called haxagonal closed packed (HCP) is constituted of a sphere on a flat surface surrounded in the same plane by six identical spheres as closely as possible . Three spheres are then placed over the first layer so that they touch each other and represent the second layer. Finally, the second layer is covered with a third layer that is identical to the bottom layer that is identical to the bottom layer in relative position. Assume radius of every sphere to be 'r'.
The empty space in this HCP unit cell is

Options

A. 0.74
B. 0.476
C. 0.32
D. 0.26

Explanation

To find the empty space in a hexagonal closed packed (HCP) unit cell, we will follow these steps:

### Step 1: Understand the structure of HCP
In an HCP structure, we have:
- Two hexagonal faces (top and bottom) and three layers of atoms.
- The first layer consists of one atom at the center and six surrounding atoms, making a total of 7 atoms in the first layer.
- The second layer consists of three atoms positioned in the gaps of the first layer.
- The third layer is identical to the first layer.

### Step 2: Calculate the volume of the unit cell
The volume of the unit cell can be calculated using the formula for the volume of a hexagonal prism:
\[ V_{cell} = \frac{3\sqrt{3}}{2} a^2 c \]
Where:
- \( a \) is the distance between the centers of two adjacent spheres in the same layer (which is \( 2r \)).
- \( c \) is the height of the unit cell, which is \( \frac{4r}{\sqrt{2}} \) for HCP.

Substituting these values:
\[ V_{cell} = \frac{3\sqrt{3}}{2} (2r)^2 \left(\frac{4r}{\sqrt{2}}\right) \]
\[ = \frac{3\sqrt{3}}{2} \cdot 4r^2 \cdot \frac{4r}{\sqrt{2}} \]
\[ = \frac{3\sqrt{3} \cdot 16r^3}{2\sqrt{2}} = 24\sqrt{3}r^3 \]

### Step 3: Calculate the volume occupied by the atoms
In HCP, there are 6 atoms in the unit cell (1 from the top face, 1 from the bottom face, and 4 from the three layers). The volume occupied by the atoms is given by:
\[ V_{atoms} = n \cdot \frac{4}{3} \pi r^3 \]
Where \( n = 6 \):
\[ V_{atoms} = 6 \cdot \frac{4}{3} \pi r^3 = 8\pi r^3 \]

### Step 4: Calculate the empty space
The empty space in the unit cell is the difference between the volume of the unit cell and the volume occupied by the atoms:
\[ V_{empty} = V_{cell} - V_{atoms} \]
Substituting the values we calculated:
\[ V_{empty} = 24\sqrt{3}r^3 - 8\pi r^3 \]

### Final Answer
The empty space in the HCP unit cell is:
\[ V_{empty} = (24\sqrt{3} - 8\pi) r^3 \]